In mathematical analysis, many real-world scenarios cannot be modeled using strict equality. Instead, they involve constraints such as "at most", "at least", "less than", or "greater than". These relationships are expressed as inequalities.
Epigraph: James Clerk Maxwell
Two real numbers or two algebraic expressions related by the symbols $<$, $>$, $\le$, or $\ge$ form an inequality.
Inequalities can be classified based on their structure, variables, and the nature of the inequality sign:
| Inequality Type | Mathematical Form / Examples | Key Characteristics |
|---|---|---|
| Numerical Inequality | $3 < 5$, $7 > 5$ | Contains only real numbers without variables. |
| Literal Inequality | $x < 5$, $y > 2$, $x \ge 3$ | Contains one or more variables. |
| Double Inequality | $3 < x < 5$, $2 < y \le 4$ | Expresses a variable bounded between two limits. |
| Strict Inequality | $ax + b < 0$, $ax + by > c$ | Uses strict relation signs ($<$ or $>$). Boundary is excluded. |
| Slack Inequality | $ax + b \le 0$, $ax + by \ge c$ | Uses weak relation signs ($\le$ or $\ge$). Boundary is included. |
| Linear Inequality in One Variable | $ax + b < 0$ (where $a \ne 0$) | Degree of the variable is exactly $1$. |
| Linear Inequality in Two Variables | $ax + by < c$ (where $a \ne 0, b \ne 0$) | Contains two variables of degree $1$. |
| Quadratic Inequality | $ax^2 + bx + c \le 0$ (where $a \ne 0$) | Degree of the variable is $2$. |
A value of the variable that makes an inequality a true statement is called a solution of the inequality. The set of all solutions is called its solution set.
To solve inequalities algebraically, we apply two primary rules that preserve the validity of the inequality:
Mathematical Problem Statement: Solve $30x < 200$ when:
(i) $x$ is a natural number,
(ii) $x$ is an integer.
Step-by-Step Algorithmic Solution Trace:
- Start with the given inequality:$$30x < 200$$
- Divide both sides by the positive number $30$ (applying Rule 2, sign remains unchanged):$$\frac{30x}{30} < \frac{200}{30}$$$$x < \frac{20}{3}$$$$x < 6.67$$
- Evaluate the solution set based on the domain constraints:
- Case (i): $x \in \mathbb{N}$ (Natural Numbers)
The natural numbers less than $6.67$ are $1, 2, 3, 4, 5, 6$.
Solution Set: $\{1, 2, 3, 4, 5, 6\}$- Case (ii): $x \in \mathbb{Z}$ (Integers)
The integers less than $6.67$ are $\dots, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6$.
Solution Set: $\{\dots, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$Final Evaluation Output: For $x \in \mathbb{N}$, solution set is $\{1, 2, 3, 4, 5, 6\}$; for $x \in \mathbb{Z}$, solution set is $\{\dots, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$.
Mathematical Problem Statement: Solve $5x - 3 < 3x + 1$ when:
(i) $x$ is an integer,
(ii) $x$ is a real number.
Step-by-Step Algorithmic Solution Trace:
- Write down the inequality:$$5x - 3 < 3x + 1$$
- Add $3$ to both sides (Rule 1):$$5x - 3 + 3 < 3x + 1 + 3$$$$5x < 3x + 4$$
- Subtract $3x$ from both sides (Rule 1):$$5x - 3x < 3x + 4 - 3x$$$$2x < 4$$
- Divide both sides by $2$ (Rule 2):$$x < 2$$
- Apply domain constraints:
- Case (i): $x \in \mathbb{Z}$
The integers less than $2$ are $\dots, -3, -2, -1, 0, 1$.
Solution Set: $\{\dots, -3, -2, -1, 0, 1\}$- Case (ii): $x \in \mathbb{R}$ (Real Numbers)
All real numbers less than $2$ are represented in interval notation.
Solution Set: $(-\infty, 2)$Final Evaluation Output: For $x \in \mathbb{R}$, the solution set is $(-\infty, 2)$.
Mathematical Problem Statement: Solve the fractional inequality:$$\frac{5 - 2x}{3} \le \frac{x}{6} - 5$$
Step-by-Step Algorithmic Solution Trace:
- Write the inequality:$$\frac{5 - 2x}{3} \le \frac{x - 30}{6}$$
- Multiply both sides by the positive LCM of denominators, which is $6$ (Rule 2, sign remains unchanged):$$2(5 - 2x) \le x - 30$$
- Expand the left-hand side:$$10 - 4x \le x - 30$$
- Subtract $x$ from both sides:$$10 - 4x - x \le -30 \implies 10 - 5x \le -30$$
- Subtract $10$ from both sides:$$-5x \le -40$$
- Divide both sides by $-5$ and reverse the inequality sign (Rule 2):$$x \ge \frac{-40}{-5}$$$$x \ge 8$$
Final Evaluation Output: The solution set for $x \in \mathbb{R}$ is $[8, \infty)$.
The solution set of a linear inequality in one variable can be visually represented on a real number line. This representation uses specific conventions to indicate whether the boundary point is included or excluded:
Mathematical Problem Statement: Solve $7x + 3 < 5x + 9$ and represent the solution graphically on a number line.
Step-by-Step Algorithmic Solution Trace:
- Simplify the inequality:$$7x - 5x < 9 - 3$$$$2x < 6$$$$x < 3$$
- Identify the boundary point: $x = 3$. Since the inequality is strict ($<$), we use an open circle at $3$.
- Shade the region to the left of $3$ (towards $-\infty$) to represent all real numbers less than $3$.
Final Evaluation Output: Solution set is $(-\infty, 3)$, represented graphically with an open circle at $3$ and a bold line extending to the left.
Mathematical Problem Statement: Solve $\frac{3x - 4}{2} \ge \frac{x + 1}{4} - 1$ and represent the solution graphically on a number line.
Step-by-Step Algorithmic Solution Trace:
- Simplify the right-hand side:$$\frac{3x - 4}{2} \ge \frac{x + 1 - 4}{4}$$$$\frac{3x - 4}{2} \ge \frac{x - 3}{4}$$
- Multiply both sides by $4$:$$2(3x - 4) \ge x - 3$$$$6x - 8 \ge x - 3$$
- Isolate the variable $x$:$$6x - x \ge -3 + 8$$$$5x \ge 5$$$$x \ge 1$$
- Identify the boundary point: $x = 1$. Since the inequality is slack ($\ge$), we use a dark circle at $1$.
- Shade the region to the right of $1$ (towards $\infty$).
Final Evaluation Output: Solution set is $[1, \infty)$, represented graphically with a dark circle at $1$ and a bold line extending to the right.
A system of linear inequalities consists of two or more inequalities in the same variable. The solution of the system is the set of all values of the variable that satisfy all inequalities in the system simultaneously (i.e., the intersection of individual solution sets).
Mathematical Problem Statement: Solve the system of inequalities:$$3x - 7 < 5 + x \quad \text{and} \quad 11 - 5x \le 1$$Represent the common solution on a number line.
Step-by-Step Algorithmic Solution Trace:
- Solve the first inequality:$$3x - 7 < 5 + x$$$$2x < 12 \implies x < 6 \quad \text{--- (1)}$$
- Solve the second inequality:$$11 - 5x \le 1$$$$-5x \le -10 \implies x \ge 2 \quad \text{--- (2)}$$
- Find the intersection of the two solution sets:$$\text{Intersection: } \{x \in \mathbb{R} : x \ge 2\} \cap \{x \in \mathbb{R} : x < 6\}$$$$2 \le x < 6$$
- Graphical representation: A dark circle at $2$ and an open circle at $6$, with a bold line connecting them.
Final Evaluation Output: The solution set of the system is $[2, 6)$.
Mathematical Problem Statement: Find all pairs of consecutive odd natural numbers, both of which are larger than $10$, such that their sum is less than $40$.
Step-by-Step Algorithmic Solution Trace:
- Let the smaller consecutive odd natural number be $x$. The next consecutive odd natural number is $x + 2$.
- Set up the system of inequalities based on the given conditions:
- Both numbers must be larger than $10$:$$x > 10 \quad \text{--- (1)}$$(Since $x > 10$, then automatically $x + 2 > 12 > 10$).
- Their sum must be less than $40$:$$x + (x + 2) < 40$$$$2x + 2 < 40$$$$2x < 38 \implies x < 19 \quad \text{--- (2)}$$
- Combine the inequalities:$$10 < x < 19$$
- Since $x$ must be an odd natural number, the possible values for $x$ are:$$x \in \{11, 13, 15, 17\}$$
- Form the pairs $(x, x + 2)$:
- For $x = 11$, the pair is $(11, 13)$.
- For $x = 13$, the pair is $(13, 15)$.
- For $x = 15$, the pair is $(15, 17)$.
- For $x = 17$, the pair is $(17, 19)$.
Final Evaluation Output: The required pairs are $(11, 13)$, $(13, 15)$, $(15, 17)$, and $(17, 19)$.
Mathematical Problem Statement: A manufacturer has $600$ litres of a $12\%$ solution of acid. How many litres of a $30\%$ acid solution must be added to it so that the acid content in the resulting mixture will be more than $15\%$ but less than $18\%$?
Step-by-Step Algorithmic Solution Trace:
- Let $x$ be the volume (in litres) of the $30\%$ acid solution to be added.
Total volume of the mixture = $(x + 600)$ litres.- Express the total acid content in the mixture:$$\text{Total Acid} = 30\% \text{ of } x + 12\% \text{ of } 600 = 0.30x + 0.12(600)$$
- Set up the double inequality representing the constraints:$$15\% \text{ of } (x + 600) < \text{Total Acid} < 18\% \text{ of } (x + 600)$$$$0.15(x + 600) < 0.30x + 72 < 0.18(x + 600)$$
- Solve the left inequality:$$0.15x + 90 < 0.30x + 72$$$$18 < 0.15x \implies x > 120 \quad \text{--- (1)}$$
- Solve the right inequality:$$0.30x + 72 < 0.18x + 108$$$$0.12x < 36 \implies x < 300 \quad \text{--- (2)}$$
- Combine the results:$$120 < x < 300$$
Final Evaluation Output: The manufacturer must add between $120$ and $300$ litres of the $30\%$ acid solution.